Question

Three capacitors, whose capacitances are 10 uF, 15 uF and 30 uF
respectively, are connected in series order, then find the equivalent
capacitance of the combination.​

Answer 1

Answer:

5 μF

Explanation:

Effective capacitance when connected in series is:

$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

Here,

$C_1 = 10$ μF

$C_2 = 15$ μF

$C_3 = 30$ μF

$\frac{1}{C_s} = \frac{1}{10} + \frac{1}{15} + \frac{1}{30}$

    =  $\frac{15+10}{150} + \frac{1}{30} = \frac{ (25 X 30) + 150}{150 X 30}$

    = $\frac{900}{4500} = \frac{1}{5}$

So, $C_s = 5$ μF

Answer 2

SOLUTION

GIVEN

Three capacitors whose capacitances are 10 μF , 15 μF and 30 μF respectively, are added in series

TO DETERMINE

The equivalent capacitance of the combination.

EVALUATION

Here it is given that three capacitors whose capacitances are 10 μF , 15 μF and 30 μF respectively, are added in series

So by the given

$\sf{C_1 = 10 \: \mu F \: , \: C_2 = 15 \: \mu F \: , \: C_3 = 30 \: \mu F}$

Let the equivalent capacitance of the combination = C

Thus we have

$\displaystyle\sf{ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} }$

$\displaystyle\sf{ \implies \: \frac{1}{C} = \frac{1}{10} + \frac{1}{15} + \frac{1}{30} }$

$\displaystyle\sf{ \implies \: \frac{1}{C} = \frac{3 + 2 + 1}{30} }$

$\displaystyle\sf{ \implies \: \frac{1}{C} = \frac{6}{30} }$

$\displaystyle\sf{ \implies \: \frac{1}{C} = \frac{1}{5} }$

$\displaystyle\sf{ \implies \: C= 5}$

FINAL ANSWER

The equivalent capacitance of the combination = 5 μF

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